高数重积分 f(x,y)=xy+1-∬D f(x,y)dσ D:x^2+y^2≦1 求∬D f(x,y)dσ=f(x,y)在D上连续
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/02 02:42:31
![高数重积分 f(x,y)=xy+1-∬D f(x,y)dσ D:x^2+y^2≦1 求∬D f(x,y)dσ=f(x,y)在D上连续](/uploads/image/z/12072354-42-4.jpg?t=%E9%AB%98%E6%95%B0%E9%87%8D%E7%A7%AF%E5%88%86+f%28x%2Cy%29%3Dxy%2B1%EF%BC%8D%26%238748%3BD+f%28x%2Cy%29d%CF%83+D%3Ax%5E2%2By%5E2%E2%89%A61+%E6%B1%82%26%238748%3BD+f%28x%2Cy%29d%CF%83%EF%BC%9Df%28x%2Cy%29%E5%9C%A8D%E4%B8%8A%E8%BF%9E%E7%BB%AD)
高数重积分 f(x,y)=xy+1-∬D f(x,y)dσ D:x^2+y^2≦1 求∬D f(x,y)dσ=f(x,y)在D上连续
高数重积分 f(x,y)=xy+1-∬D f(x,y)dσ D:x^2+y^2≦1 求∬D f(x,y)dσ=
f(x,y)在D上连续
高数重积分 f(x,y)=xy+1-∬D f(x,y)dσ D:x^2+y^2≦1 求∬D f(x,y)dσ=f(x,y)在D上连续
设Z=∬D f(x,y)dσ,对原式两边在D平面上求而重积分:
∬D f(x,y)dσ=∬D [xy+1-∬D f(x,y)dσ]dσ
就是:Z=∬D (xy+1-Z)dσ
因为积分区域是个半径为1的圆,关于x轴和y轴都对称,所以积分xy在其上结果是0
那么Z=(1-Z)*π
所以Z=π/(1+π)
也就是∬D f(x,y)dσ=π/(1+π)